Two identical conducting spheres $A$ and $B$, carry equal charge. They are separated by a distance much larger than their diameters, and the force between them is $F$. A third identical conducting sphere, $C$, is uncharged. Sphere $C$ is first touched to $A$, then to $B$, and then removed. As a result, the force between $A$ and $B$ would be equal to :
To solve this problem, we need to understand the process of sharing charge between conducting spheres due to touching and then calculate the resultant force between the two initially charged spheres.
Initially, spheres A and B have equal charges q each, and they are separated by a distance, resulting in an electrostatic force F. By Coulomb's Law, this force is given by:
F = k \frac{q^2}{r^2},
where k is Coulomb's constant and r is the distance between A and B.
Sphere C is initially uncharged and then first touched with sphere A. When they are touched, the charge is redistributed equally between A and C because they are identical. Thus each gets a charge of:
\frac{q}{2}.
Sphere C now with charge \frac{q}{2} is then touched to sphere B, and again the charges redistribute equally. B initially had charge q, hence after touching, both B and C will have:
\frac{q + \frac{q}{2}}{2} = \frac{3q}{4}.
Sphere C is then removed. The final charge on sphere A is \frac{q}{2} and on B is \frac{3q}{4}.
The electrostatic force F' between spheres A and B is now calculated using Coulomb's Law:
F' = k \frac{\left(\frac{q}{2}\right)\left(\frac{3q}{4}\right)}{r^2} = k \frac{3q^2}{8r^2}.
Simplifying, we compare this with the initial force:
F = k \frac{q^2}{r^2}.
The new force F':
F' = \frac{3}{8}F.
Hence, the correct answer is \frac{3F}{8}.
Thus, after considering the redistribution of charge through the touching of sphere C to both A and B, the force between A and B decreases to \frac{3F}{8}.
